Question

An impulse $\text{J}$ is applied on a ring of mass $\text{m}$ along a line passing through its centre $\text{O}$ . The ring is placed on a rough horizontal surface. The linear velocity of centre of ring once it starts rolling without slipping is

• A. $\frac{\text{J}}{\text{m}}$
• B. $\frac{\text{J}}{\text{2m}}$
• C. $\frac{\text{J}}{\text{4m}}$
• D. $\frac{\text{J}}{\text{3m}}$

Answer 1

Answer: (B)

Let $\text{v}$ be the velocity of COM of the ring just after the impulse is applied and $\text{v’}$ is the velocity when pure rolling starts.
Angular velocity $\text{ω}$ of the ring at this instant will be $\omega =\frac{\text{v’}}{\text{r}}$
$\Rightarrow$ Impulse = change in linear momentum
we have, $\text{J}=\text{mv}$
or $\text{v=J/m}$
Between the two positions shown in figure force of
friction on the ring acts backwards. Angular momentum
of the ring about bottommost point will remain conserved
$\therefore {\text{L}}_{\text{i}}={\text{L}}_{\text{f}}$
or $\text{mvr}=\text{mv’r}+\text{l}\omega$
$=\text{mv’ r +}\left({\left(\text{mr}\right)}^2\right)\frac{\text{v’}}{\text{r}}=2\text{mv’ r}$
$\therefore$ $\text{v’}=\frac{v}{2}=\text{J/2m}$